You know that $\sin^2+\cos^2=1$. As $\sin(24)=p$ you have $p^2 + \cos^2(24)=1$. Solve this equation for $\cos(24)$.
For the second part you can use some trigonometric theorems.
For example
$$ \sin(\alpha) \cdot \sin(\beta) = -\frac{1}{2} (\cos(\alpha+\beta) -\cos(\alpha - \beta))$$

This gives you
$$\sin(168^\circ) \cdot \sin(-78^\circ)=-\frac{1}{2} ( \cos(90^\circ)- \cos(246^\circ))$$
We know that $\cos(90^\circ)=0$ and that $\cos(x+180^\circ)=-\cos(x)$ hence we have
$$-\frac{1}{2} \cos(66^\circ)=-\frac{1}{2}\cos(90^\circ-24^\circ)=-\frac{1}{2} \sin(24^\circ)$$